A Naval Architect that does some contract work for the place I work at has done some calcs for expected wave heights in enclosed waters to ensure floating pontoons, marinas etc can cope with expected worse case scenarios. He was using a reference book that was published either by the US Navy or US Army Engineers. I'll have a chat with him at work on Monday to see if I get you any further info for you.

I am trying to calculate salt water wave heights inside a Bay given a known distance in miles and a wind scale up to Hurricane strength.

No Ocean Swell or currents are factored into this calculation and depth can be assumed to be more than 90 feet.

I know that formula exists, just can’t find it…… any ideas??

I looked around a bit in my books and on the net.
This link Fetch- and Depth-Limited Wave Height and Period is probably your best bet. It is not the formula itself, it's even better, a calculator for what you are asking for.
Check the "Disclaimer" for info about formulas that this Java-applet is based on.
Bowditch has a table, http://www.irbs.com/bowditch/pdf/chapt33.pdf ( page 2), the shortest fetch is however as much as 10 miles.
I have briefly compared these two methods, and Bowditch gives a slightly higher value.
I hope this will help you!

Location: [S]Hamble (Spring and Fall)[/S], Cowes (Winter), Baltic (Summer) (the boat!); somewhere in the air (me!)

Boat: Cutter-Rigged Moody 54

Posts: 16,798

Quote:

Originally Posted by cagney

I looked around a bit in my books and on the net.
This link Fetch- and Depth-Limited Wave Height and Period is probably your best bet. It is not the formula itself, it's even better, a calculator for what you are asking for.
Check the "Disclaimer" for info about formulas that this Java-applet is based on.
Bowditch has a table, http://www.irbs.com/bowditch/pdf/chapt33.pdf ( page 2), the shortest fetch is however as much as 10 miles.
I have briefly compared these two methods, and Bowditch gives a slightly higher value.
I hope this will help you!

Very interesting. This gives much more modest results than "20 knots = 20 feet". 20 meters/second (about 40 knots), over 200km and infinite depth, gives 6.3 meters (a little over 20 feet), according to the calculator.

Over 2000km, the same conditions give a terrifying 16.4 meters.

10 meters/second (20 knots) over the same 2,000km gives only 4 meters.

I guess it would be rare to have 40 knot winds acting over 2,000km of open ocean.

Very interesting. This gives much more modest results than "20 knots = 20 feet". 20 meters/second (about 40 knots), over 200km and infinite depth, gives 6.3 meters (a little over 20 feet), according to the calculator.

Over 2000km, the same conditions give a terrifying 16.4 meters.

10 meters/second (20 knots) over the same 2,000km gives only 4 meters.

I guess it would be rare to have 40 knot winds acting over 2,000km of open ocean.

Interesting.

As a sanity check, I put in 8 m/s, which is about 15 knots, 2000 km fetch, and 2000 m depth, essentially infinite. This should replicate conditions on the north shore of the caribbean islands during steady state trade winds that blow over long fetches of water at fairly steady speeds. I got 2.4 m wave height, about 7 feet. Seems about right, maybe a foot or so too high, but it's rare to get steady 15 knots and all from exactly the same direction. From this one case, I think this calculator does pass a crude sanity check.

__________________The pessimist complains about the wind; the optimist expects it to change; the realist adjusts the sails.
- William Arthur Ward

Have a look at todays wind speed / wave height graphic for the next few days.
Its interesting because the wind sppeds are quite stable, but increasing, 15, 20 then 30. Look at the corresponding wave heights.

The fetch in this area is short and I don't know how their model works with it. Perhaps its not able to calculate fetch for this one port.

As a sanity check, I put in 8 m/s, which is about 15 knots, 2000 km fetch, and 2000 m depth, essentially infinite. This should replicate conditions on the north shore of the caribbean islands during steady state trade winds that blow over long fetches of water at fairly steady speeds. I got 2.4 m wave height, about 7 feet. Seems about right, maybe a foot or so too high, but it's rare to get steady 15 knots and all from exactly the same direction. From this one case, I think this calculator does pass a crude sanity check.

Remember the current the trade wind generates, this is decreasing the wave height, so a calculated value showing a higher value than you observe in the Carribbean is to be expected.

In his book "Modern Marine Weather" David Burch states that you can count on the wind driven current to be in the region of 3% of the wind strength after at least half a days steady blow.

Excellent point, cagney. So Burch's model would predict about a half knot of current, which seems about right, and that should lower the predicted wave height. I looked for a formula that relates current to wave height but couldn't find one. But at least the current correction is working in the correct direction for the wave height calculator.

__________________

__________________The pessimist complains about the wind; the optimist expects it to change; the realist adjusts the sails.
- William Arthur Ward